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40 CHAPTER 1. ORIGINS OF QUANTUM PHYSICS

-

6

x0

O0x2

x

-

6

k

k0 k0

Mk2

Figure 1.13 Two localized wave packets: O0x 2Ha214ex2a2 eik0x and Mk a22H14ea

2kk024; they peak at x 0 and k k0, respectively, and vanish far away.

the particle between x and xdx . What about the physical interpretation ofMk? From (1.95)and (1.96) it follows that

=**

O0x2dx =**

Mk2dk (1.97)

then ifOx is normalized so is Mk, and vice versa. Thus, the function Mk can be interpreted

most naturally, like O0x, as a probability amplitude for measuring a wave vectork for a parti-

cle in the state Mk. Moreover, while Mk2 represents the probability density for measuring kas the particles wave vector, the quantity Pk dk Mk2dk gives the probability offindingthe particles wave vector between k and k dk.

We can extract information about the particles motion by simply expressing its correspond-

ing matter wave in terms of the particles energy, E, and momentum, p. Using k ph,dk dph, E h and redefining Mp Mk

Th, we can rewrite (1.94) to (1.96) as

follows:

Ox t 1T2H

h

=*

*

MpeipxEth dp (1.98)

O0x 1T

2H h

=**

Mpei px hdp (1.99)

Mp 1T2H h

=**

O0xei px h dx (1.100)

where Ep is the total energy of the particle described by the wave packet Ox t and Mp isthe momentum amplitude of the packet.

In what follows we are going to illustrate the basic ideas of wave packets on a simple,

instructive example: the Gaussian and square wave packets.

Example 1.8 (Gaussian and square wave packets)

(a) Find Ox 0 for a Gaussian wave packet Mk A exp da2k k024e, where A isa normalization factor to be found. Calculate the probability offinding the particle in the region

a2 n x n a2.

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1.8. WAVE PACKETS 41

(b) Find Mk for a square wave packet O0x |

Aeik0x x n a0 x a

Find the factor A so that Ox is normalized.

Solution

(a) The normalization factor A is easy to obtain:

1 =**

Mk2dk A2=**

expva2

2k k02

wdk (1.101)

which, by using a change of variable z k k0 and using the integral5** e

a2z22dz T

2Ha, leads at once to A T

aT

2H [a22H]14. Now, the wave packet correspondingto

Mk t

a2

2H

u14exp

va

2

4k k02

w(1.102)

is

O0x 1T2H

=**

Mkeik x dk 1T2H

ta2

2H

u14 =**

ea2kk024ik x dk (1.103)

To carry out the integration, we need simply to rearrange the exponents argument as follows:

a2

4k k02 ik x

va

2k k0

i x

a

w2 x

2

a2 ik0x (1.104)

The introduction of a new variable y ak k02 i xa yields dk 2dya, and whencombined with (1.103) and (1.104), this leads to

O0x 1T2H

ta2

2H

u14 =**

ex2a2 eik0x ey

2

t2

ady

u

1TH

t2

Ha2

u14ex

2a2 eik0x=**

ey2

dy (1.105)

Since5** e

y2 dy TH , this expression becomes

O0x t

2

Ha2

u14ex

2a2 eik0x (1.106)

where eik0x is the phase ofO0x; O0x is an oscillating wave with wave numberk0 modulated

by a Gaussian envelope centered at the origin. We will see later that the phase factoreik0x has

real physical significance. The wave function O0x is complex, as necessitated by quantum

mechanics. Note that O0x, like Mk, is normalized. Moreover, equations (1.102) and (1.106)

show that the Fourier transform of a Gaussian wave packet is also a Gaussian wave packet.

The probability offinding the particle in the region a2 n x n a2 can be obtained atonce from (1.106):

P=a2a2

O0x2dx U

2

Ha2

=a2a2

e2x2a2 dx 1T

2H

=11

ez22dz 2

3 (1.107)

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42 CHAPTER 1. ORIGINS OF QUANTUM PHYSICS

where we have used the change of variable z 2xa.(b) The normalization ofO0x is straightforward:

1 =**

O0x2dx A2=aa

eik0x eik0x dx A2=aa

dx 2aA2 (1.108)

hence A 1T2a. The Fourier transform ofO0x is

Mk 1T2H

=**

O0xei kx d x 1

2T

Ha

=aa

eik0x eik x dx 1THa

sin [k k0a]k k0

(1.109)

1.8.2 Wave Packets and the Uncertainty Relations

We want to show here that the width of a wave packet O0x and the width of its amplitude

Mk are not independent; they are correlated by a reciprocal relationship. As it turns out, the

reciprocal relationship between the widths in the x and k spaces has a direct connection to

Heisenbergs uncertainty relation.

For simplicity, let us illustrate the main ideas on the Gaussian wave packet treated in theprevious example (see (1.102) and (1.106)):

O0x t

2

Ha2

u14ex

2a2 eik0x Mk t

a2

2H

u14ea

2kk024 (1.110)

As displayed in Figure 1.13, O0x2 and Mk2 are centered at x 0 and k k0, respec-tively. It is convenient to define the half-widths x and kas corresponding to the half-maxima

ofO0x2 and Mk2. In this way, when x varies from 0 to x and k from k0 to k0 k,the functions O0x 2 and Mk2 drop to e12:

Ox 02

O0 0

2

e12 Mk0 k2

Mk0

2

e12 (1.111)

These equations, combined with (1.110), lead to e2x2a2 e12 and ea2k22 e12,

respectively, or to

x a2

k 1a (1.112)

hence

xk 12

(1.113)

Since k ph we havexp h

2 (1.114)

This relation shows that if the packets width is narrow in x-space, its width in momentum

space must be very broad, and vice versa.A comparison of (1.114) with Heisenbergs uncertainty relations (1.57) reveals that the

Gaussian wave packet yields an equality, not an inequality relation. In fact, equation (1.114) is